npleted 25 out of 36 Submit on 8 of 36 A sample of an ideal gas has a volume of 3.80 L at 12.60 °C and 1.50 atm. What is the volume of the gas at 18.80 °C and 0.987 atm? V = L
Ideal and Real Gases
Ideal gases obey conditions of the general gas laws under all states of pressure and temperature. Ideal gases are also named perfect gases. The attributes of ideal gases are as follows,
Gas Laws
Gas laws describe the ways in which volume, temperature, pressure, and other conditions correlate when matter is in a gaseous state. The very first observations about the physical properties of gases was made by Robert Boyle in 1662. Later discoveries were made by Charles, Gay-Lussac, Avogadro, and others. Eventually, these observations were combined to produce the ideal gas law.
Gaseous State
It is well known that matter exists in different forms in our surroundings. There are five known states of matter, such as solids, gases, liquids, plasma and Bose-Einstein condensate. The last two are known newly in the recent days. Thus, the detailed forms of matter studied are solids, gases and liquids. The best example of a substance that is present in different states is water. It is solid ice, gaseous vapor or steam and liquid water depending on the temperature and pressure conditions. This is due to the difference in the intermolecular forces and distances. The occurrence of three different phases is due to the difference in the two major forces, the force which tends to tightly hold molecules i.e., forces of attraction and the disruptive forces obtained from the thermal energy of molecules.
![**Ideal Gas Law Problem: Determining the Volume at Different Conditions**
**Problem Statement:**
A sample of an ideal gas has a volume of 3.80 L at 12.60°C and 1.50 atm. What is the volume of the gas at 18.80°C and 0.987 atm?
\[ V = \_\_\_\_\_\_\_\_\_ \text{L} \]
**Explanation:**
This problem deals with the properties of an ideal gas and how its volume changes when the temperature and pressure change. To solve this, we can use the combined gas law, which is formulated as:
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]
Where:
- \( P_1 \), \( V_1 \), and \( T_1 \) are the initial pressure, volume, and temperature.
- \( P_2 \), \( V_2 \), and \( T_2 \) are the final pressure, volume, and temperature.
**Given Data:**
- Initial Volume \( V_1 = 3.80 \text{ L} \)
- Initial Temperature \( T_1 = 12.60^\circ\text{C} = 12.60 + 273.15 = 285.75 \text{ K} \)
- Initial Pressure \( P_1 = 1.50 \text{ atm} \)
- Final Temperature \( T_2 = 18.80^\circ\text{C} = 18.80 + 273.15 = 291.95 \text{ K} \)
- Final Pressure \( P_2 = 0.987 \text{ atm} \)
**Objective:**
- Determine the final volume \( V_2 \).
This involves finding the volume \( V \) when the gas is at the new specified conditions of temperature and pressure using the formula provided.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fea81c1af-6d79-4288-acfc-85b1019cce6f%2F791d0e15-313c-48ab-b6ad-4c0820c32e26%2Fwnt3zx_processed.jpeg&w=3840&q=75)

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